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प्रश्न
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उत्तर
'An electric bulb is rated 250 W-230V'; means that if the bulb is lighted on a 230V supply it consumes 25 ow electrical power or 25 OJ of electrical energy converts into heat and light in 1 second.
The safe Ii m it of current through the bulb is:
I = `"P"/"V" = 250/230` = 1.1 A
Current through a 60W lamp rated for 250V is:
I = `"P"/"V" = 60/250` = 0.24 A
We know that resistance of an appliance remains constant.
Resistance of the 1 amp =
R = `"P"/"I"^2 = 60/(0.24)^2 Omega`
If the line voltage falls to 200 V, power =
P = `"V"^2/"R" = (200 xx 200 xx (0.24)^2)/60` = 38.4 watt
Thus power of the lamp reduces to 38. 4 watt.
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संबंधित प्रश्न
What is an electric current?
In the circuit diagram given below five resistances of 10 Ω, 40 Ω, 30 Ω, 20 Ω, and 60 Ω are connected as shown to a 12 V battery.
Calculate:
(a) total resistance in the circuit.
(b) total current flowing in the circuit.
The following table shows the current in Amperes and potential differences in Volts.
- Find the average resistance.
- What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
- Which law will the graph prove? Explain the law.
| V (volts) | I (amp) |
| 4 | 9 |
| 5 | 11.25 |
| 6 | 13.5 |
Find the effective resistance in the following circuit diagrams (Fig.):
Complete the following table. The first answer has been given as an example
| Quantity | Unit |
|
(a) Electrical potential Volt ( b) Resistance ( c) Power ( d) Energy ( e) Resistivity |
Volt ........ .......... ............ .............. |
A battery of e.m.f. 15 V and internal resistance 2 Ω is connected to two resistors of resistances 4 ohm and 6 ohm joined (a) in series. Find in each case the electrical energy spent per minute in 6 Ω resistor.
Fill in the blank :
1 mA = ............................. A.
For current to flow, one needs art open circuit.
